Variation of Parameters — Question 1

Question 1 – Solution

Strategy. Divide by the leading coefficient before solving for the parameter derivatives; check the answer in the original equation.

Step 1: Normalize and derive. The normalized forcing is $g(t)=t$, and $W=t(2t)-t^2=t^2\ne 0$. With $y=u_1t+u_2t^2$, impose $tu_1'+t^2u_2'=0$. Differentiating after this cancellation gives $$tu_1'+t^2u_2'=0,\qquad u_1'+2tu_2'=t.$$ Thus $u_1'=-t$ and $u_2'=1$.

Step 2: Integrate and fit. Choose $u_1=-t^2/2$, $u_2=t$; constants of integration contribute only homogeneous terms. Hence $y_p=t^3/2$ and $$y=Ct+Dt^2+t^3/2.$$ The data give $C+D=-1/2$ and $C+2D=-3/2$, so $C=1/2$, $D=-1$. Therefore $$\boxed{y=\tfrac 12t(t-1)^2.}$$

Step 3: Verify the original equation. For the operator $L[y]=t^2y''-2ty'+2y$, one has $L[t]=L[t^2]=0$ and $L[t^3]=2t^3$. Thus $L[y]=t^3$. Also $y(1)=0$ and $y'=(3t^2-4t+1)/2$, so $y'(1)=0$. The normalized coefficients are continuous on $(0,\infty)$, giving uniqueness there.

Step 4: Diagnose the missing division. Using $g=t^3$ instead gives $u_1'=-t^3$, $u_2'=t^2$ and $$\widetilde y_p=-t^5/4+t^5/3=t^5/12.$$ Since $L[t^5]=(20-10+2)t^5=12t^5$, this expression satisfies $L[\widetilde y_p]=t^5$, not $t^3$. Adding homogeneous terms cannot repair its forcing. The parameter formulas apply to the normalized equation; omitting the division multiplies the intended original forcing by $t^2$.